Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
	int x=0,f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline int read(){
	int x=0,f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e5;
struct S1{
	#define ls (p<<1)
	#define rs (p<<1|1)
	int tree[(N<<3)+10];
	void pushdown(int p){
		if (!tree[p])	return;
		tree[ls]=tree[rs]=tree[p];
		tree[p]=0;
	}
	void Modify(int p,int l,int r,int x,int y,int v){
		if (x>y)	return;
		if (x<=l&&r<=y){
			tree[p]=v;
			return;
		}
		pushdown(p);
		int mid=(l+r)>>1;
		if (x<=mid)	Modify(ls,l,mid,x,y,v);
		if (y>mid)	Modify(rs,mid+1,r,x,y,v);
	}
	int Query(int p,int l,int r,int x){
		if (l==r)	return tree[p];
		pushdown(p);
		int mid=(l+r)>>1;
		if (x<=mid)	return Query(ls,l,mid,x);
		else	return Query(rs,mid+1,r,x);
	}
	#undef ls
	#undef rs
}ST;//Segment Tree
int A[N+10],B[N+10];
int L[N+10],R[N+10];
ll sum[N+10],f[N+10];
int list[(N<<1)+10];
int n,K,T,tot;
ll Get(int x){
	int tmp=ST.Query(1,1,T,x);
	if (!tmp)	return 0;
	return f[tmp]+(list[L[tmp]]-list[x]+K)%K;
}
int main(){
	n=read(),K=read();
	bool flag=0;
	for (int i=1;i<=n;i++){
		A[i]=read(),B[i]=read();
		sum[i]=sum[i-1]+A[i];
		if (B[i]==2)	continue;
		if (2*A[i]>K)	flag=1;
	}
	if (flag){
		printf("-1\n");
		return 0;
	}
	for (int i=1;i<=n;i++){
		if (B[i]==1){
			L[i]=(-2*sum[i-1]%K+K)%K;
			R[i]=(-2*sum[i  ]%K+K)%K;
		}else	L[i]=0,R[i]=K-1;
		list[++tot]=L[i];
		list[++tot]=R[i];
	}
	sort(list+1,list+1+tot);
	T=unique(list+1,list+1+tot)-list-1;
	for (int i=1;i<=n;i++){
		L[i]=lower_bound(list+1,list+1+T,L[i])-list;
		R[i]=lower_bound(list+1,list+1+T,R[i])-list;
	}
	for (int i=n;i;i--){
		f[i]=Get(L[i]);
		if (R[i]<L[i])	ST.Modify(1,1,T,R[i]+1,L[i]-1,i);
		else	ST.Modify(1,1,T,1,L[i]-1,i),ST.Modify(1,1,T,R[i]+1,T,i);
	}
	ll Ans=f[1];
	for (int i=1;i<=T;i++)	Ans=min(Ans,Get(i));
	printf("%lld\n",Ans+2*sum[n]);
	return 0;
}